# Giant Swing, Uniform Circular Motion

1. Feb 12, 2008

### clope023

[SOLVED] Giant Swing, Uniform Circular Motion

1. The problem statement, all variables and given/known data

The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.00 long, the upper end of the cable being fastened to the arm at a point 3.00 from the central shaft.

A) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of with the vertical.

B) Does the angle depend on the weight of the passenger for a given rate of revolution?

2. Relevant equations

R = Lsin(\theta)

v = $$\sqrt{gtan(\theta)R}$$

T = 2$$\Pi$$R/v

3. The attempt at a solution

tried using L = 3+5m*sin($$\theta$$)) to get 4m

R then equals = 2m

v then equals = $$\sqrt{9.8*2*tan\theta}$$ = 11.3m/s

T then equals 2$$\Pi$$(2m)/11.3m/s = 1.1s wrong

I tried a few other combinations where I used 3+(5sintheta) as the L and got T=4.4s which was wrong as well.

I'm not sure what I'm doing wrong, I think I have the correct equations and I know I have the right angle and distances, I guess I'm just not sure how to derive the right length and and radius.

I'm also thinking that for part B the weight will determine the angle that the seat swings, but I don't want to risk losing the only chance I have on that part of the problem (masteringphysics).

any and all help is greatly appreciated.

Last edited: Feb 12, 2008
2. Feb 12, 2008

### lavalamp

You might want to go back and look at this bit again:
sin(30) = 0.5, so 5*0.5 + 3 = 5.5m, and you don't need to divide it by 2, because that is the radius, not the diameter.

3. Feb 12, 2008

### PhanthomJay

It appears that you are confusing L with R. If L is the length of the cable, then what is R, which is the radius measured from the passenger to the central shaft? Also, if you draw a free body diagram and use Newton's laws, it will help to see whether or not the the mass comes into play. You shouldn't blindly be using a formula..

4. Feb 12, 2008

### clope023

I wrote it before, R = Lsin(30)

I was pretty sure it was the length of the cable alone (which would've been 5) I just wasn't sure where the distance of the angle of the seat from the vertical shaft (3m) played in.

if it's only length of cable alone, then R = 5(sin30) so R = 2.5m, does that seem more correct?

on the angle's dependance on the weight, I've drawn a free body diagram and it makes the sum of the forces be

(sigma)Fx = -marad + mgcos30 = 0

(sigma)Fy = T - mgsin30 = 0

so I would imagine the angle is dependant on the weight of the seat.

5. Feb 12, 2008

### lavalamp

6. Feb 12, 2008

### clope023

thanks, I got the answer to the first part using that radius, I wasn't picturing in my head that the swing literally makes a circle when viewed overhead and the distance of 3m + (5sin30) was the radius of that circle.

now fro for the 2nd part, as I said I did a free body diagram and found the forces acting in each direction, but I'm actually still not certain, I believe the mass would cancel out from the 2 equations if I set them equal to each other, so the answer would be that it doesn't depend on the mass?

7. Feb 12, 2008

### PhanthomJay

Your FBD's are not correct. In the y direction, mg acts straight down, and Tcos30 acts straight up. In the x direction, only Tsin30 acts.

8. Feb 12, 2008

### clope023

so then in that case Tcos30 = mg

so the angle is proportional to the mass?

9. Feb 12, 2008

### clope023

scracth that the angle is not proportionate to the mass, thanks for the help.

10. Feb 28, 2008

### nightlaei71

I have the same problem as well, and I don't quite understand how velocity was derived.